16.9.2 The Doppler Effect for Light Waves

The Doppler effect is observed for all types of waves, not just sound waves. If a source of light waves is receding from you, the wavelength $λ_{-}$ that you detect is longer than the wavelength $λ_{0}$ emitted by the source.

Although the reason for the Doppler shift for light is the same as for sound waves, there is one fundamental difference. We derived Equations 16.65 for the Doppler-shifted frequencies by measuring the wave speed v relative to the medium. For electromagnetic waves in empty space, there is no medium. Consequently, we need to turn to Einstein’s theory of relativity to determine the frequency of light waves from a moving source. The result, which we state without proof, is

\begin{array}{l} λ_{-} = \sqrt{\frac{1 + v_{s} / c}{1 - v_{s} / c}} λ_{0} (receding source) \\ λ_{+} = \sqrt{\frac{1 - v_{s} / c}{1 + v_{s} / c}} λ_{0} (approaching source) \end{array}

(16.67)

Here $v_{s}$ is the speed of the source relative to the observer.

The light waves from a receding source are shifted to longer wavelengths $(λ_{-} > λ_{0})$ . Because the longest visible wavelengths are perceived as the color red, the light from a receding source is red shifted. That is not to say that the light is red, simply that its wavelength is shifted toward the red end of the spectrum. If $λ_{0} = 470 nm$ (blue) light emitted by a rapidly receding source is detected at $λ_{-} = 520 nm$ (green), we would say that the light has been red shifted. Similarly, light from an approaching source is blue shifted, meaning that the detected wavelengths are shorter than the emitted wavelengths $(λ_{+} < λ_{0})$ and thus are shifted toward the blue end of the spectrum.

EXAMPLE 16.12 Measuring the velocity of a galaxy

Hydrogen atoms in the laboratory emit red light with wavelength 656 nm. In the light from a distant galaxy, this “spectral line” is observed at 691 nm. What is the speed of this galaxy relative to the earth?

MODEL The observed wavelength is longer than the wavelength emitted by atoms at rest with respect to the observer (i.e., red shifted), so we are looking at light emitted from a galaxy that is receding from us.

SOLVE Squaring the expression for $λ_{-}$ in Equations 16.67 and solving for $v_{s}$ give

\begin{matrix} v_{s} & = & \frac{{(λ_{-} / λ_{0})}^{2} - 1}{{(λ_{-} / λ_{0})}^{2} + 1} c \\ = & \frac{(691 {nm/656 nm)}^{2} - 1}{(691 {nm/656 nm)}^{2} + 1} c \\ = & 0.052 c = 1.56 \times 10^{7} m / s \end{matrix}

REVIEW The galaxy is moving away from the earth at about 5% of the speed of light!

In the 1920s, an analysis of the red shifts of many galaxies led the astronomer Edwin Hubble to the conclusion that the galaxies of the universe are all moving apart from each other. Extrapolating backward in time must bring us to a point when all the matter of the universe—and even space itself, according to the theory of relativity—began rushing out of a primordial fireball. Many observations and measurements since have given support to the idea that the universe began in a Big Bang about 14 billion years ago.

As an example, FIGURE 16.31 is a Hubble Space Telescope picture of a quasar, short for quasistellar object. Quasars are extraordinarily powerful sources of light and radio waves. The light reaching us from quasars is highly red shifted, corresponding in some cases to objects that are moving away from us at greater than 90% of the speed of light. Astronomers have determined that some quasars are 10 to 12 billion light years away from the earth, hence the light we see was emitted when the universe was only about 25% of its present age. Today, the red shifts of distant quasars and supernovae (exploding stars) are being used to refine our understanding of the structure and evolution of the universe.

A photograph of Hubble Space Telescope picture of a quasar.

CHAPTER 16 CHALLENGE EXAMPLE Decreasing the sound

The loudspeaker on a homecoming float—mounted on a pole—is stuck playing an annoying 210 Hz tone. When the speaker is 10 m away, you measure the sound to be a loud 95 dB at 208 Hz. How long will it take for the sound intensity level to drop to a tolerable 55 dB?

MODEL The source is on a pole, so model the sound waves as uniform spherical waves. Assume a temperature of $20 °C$ .

SOLVE The 208 Hz frequency you measure is less than the 210 Hz frequency that was emitted, so the float must be moving away from you. The Doppler effect for a receding source is

f_{-} = \frac{f_{0}}{1 + v_{s} / v}

We can solve this to find the speed of the float:

v_{s} = (\frac{f_{0}}{f_{-}} - 1) v = (\frac{210 Hz}{208 Hz} - 1) \times 343 m/s = 3.3 m/s

The sound intensity of a spherical wave decreases with the inverse square of the distance from the source. A sound intensity level $β$ corresponds to an intensity $I = I_{0} \times 10^{β / 10 dB}$ . where $I_{0} = 1.0 \times 10^{- 12} W / m^{2}$ . At the initial 95 dB, the intensity is

I_{1} = I_{0} \times 10^{9.5} = 3.2 \times 10^{- 3} W / m^{2}

At the desired 55 dB, the intensity will have dropped to

I_{2} = I_{0} \times 10^{5.5} = 3.2 \times 10^{- 7} W / m^{2}

The intensity ratio is related to the distances by

\frac{I_{1}}{I_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}

Thus the sound will have dropped to 55 dB when the distance to the speaker is

r_{2} = \sqrt{\frac{I_{1}}{I_{2}}} r_{1} = \sqrt{10^{4}} \times 10 m = 1000 m

The float has to travel $Δ x = 990 m,$ which will take

Δ t = \frac{Δ x}{v_{s}} = \frac{990 m}{3.3 m / s} = 300 s = 5.0 min

REVIEW To drop the sound intensity level by 40 dB requires decreasing the intensity by a factor of $10^{4}$ . And with the intensity depending on the inverse square of the distance, that requires increasing the distance by a factor of 100. Floats don’t move very fast—3.3 m/s is about 7 mph—so needing several minutes to travel the $\approx 1000 m$ seems reasonable.